Answer Happy

Turning our attention to B, we see that the probability that B wins in the first round is Pr(A misses) x Pr(B hits) = (1-1)× As with A, we continue to find that the probability that B wins in round i isq; hence, summing over i, the probability that B wins is also. Finally, by subtraction, the probability that C wins is. (Alternatively, this result may be obtained using arguments similar to those above: the probability that C wins in round i is found to be the same as for the other two players.) (b) From the results obtained in solving part (a), we see that the number of throws 7 in the game has the probability function Hence Pr(T = i) = E(T) = (1+2+3+ (4+5+6)q + (7 + 8 + 9)q² + = 3 (2+5g +8q² + ...) =(3+69 + 9q² + ) - (1 + q + q² + ) 10 (1 (1- 1 i = 1, 2, 3, 10' 19, i = 4, 5, 6, 19², i = 7, 8, 9, etc. 10 9)² 1- 9 10 1 - 9 3 10 7)² 1- 10 where p = 1-9. 1 expansion of (1-q) 7 10 = 101 = 9. } Notes (1) We are, throughout our solution, assuming that the results of different throws are independent. (2) Some readers may wonder how we arrived at our solution to part (b). In finding the expectation and variance of a discrete probability distribution one frequently finds that the series to be summed are closely related to the series which one encounters in demonstrating that the probability function of the distribution sums to 1. Thus, for example, we encounter binomial series when dealing with the binomial distribution, exponential series when dealing with the Poisson distribution, and, in particular, binomial expansions of negative powers of 1-q, where q = 1-p, when dealing with the geometric distribution (see Problem 2A.10). Some links between the present distribution and the geometric distribution are explored briefly in Note 5. To demonstrate that the probability function of 7 sums to 1, we note that the sum is p(1 + q + q² +9³ + ), Since the series in brackets can be recognised as the binomial it is possible that, in seeking E(7), we may find other binomial

expansions of 1- q useful; in particular, we recollect that (1-q) 2 = 1 + 2q + 3q² + 49³ + The series we require to evaluate in finding E(T) has coefficients (2, 5, 8,) in arithmetic progression, with a common difference of 3, and the expansion of 3(1-q) 2 has the same property. Thus the difference between these series must be a series with all coefficients equal, i.e. a multiple of (1-q) ¹. An alternative approach, which some may prefer, is the following. The series to be summed is S = 2 + 5g +8q² + 11q³+.... (then E(T) =S). Noting that qS = 2q + ·59² +8q³+... and subtracting, we obtain (19)S 2 + 3q + 3q² + 3q³ + We may now recognise that the right hand side of this equation is equal to 2 + 3q (1-q)-¹, and hence proceed to evaluate S. Alternatively, we may repeat the process of multiplying by q and subtracting, obtaining (1-q)²S = 2 + q. Substituting the value for q, we obtain S = 30, and hence E(T) = 9. (3) There is another way of solving part (a). Since the argument is similar to that described in Note 2 to Problem 1A.6, we shall not give full details here. The basis of the method is that we condition on the fact that the game finishes in round k (k = 1, 2, ...). If the game has not ended before round k, the probabilities that A, B and C will win in this round are all Thus the conditional probabilities that they will win, given that the game ends in this round, are all equal to; since these do not depend on k, they are also the unconditional probabilities. (4) Another way of solving the problem is through the use of recurrence relationships. As our first illustration of this approach, we shall find the probability that A wins call this PA. Then PA Pr(A wins in round 1) + Pr(A wins in a subsequent round) 10 + Pr(all three miss in round 1) x Pr(A wins in a subsequent round all three miss in round 1) 10+ TOPA We have now obtained a recurrence relationship for PA which, on rearrangement, gives the equation PA10, with the solution Pa = ₁ А A similar approach is also possible with part (b). If we let E denote the expected number of throws in the game, then E = = (×1) + (×2) + (×3) + ( ×(3 + E)). (This follows from the fact that the number of throws in the game takes each of the values 1, 2 and 3 with probability, and otherwise, with probability, the game effectively 'restarts' after the first three throws.) The above recurrence relationship simplifies toE=, leading to E = 9. (5) In solving part (b) we could have made use of the fact that the number of rounds in the game (including the final, possibly incomplete, round) has a geometric distribution with parameter p = The expected number of rounds in the game is therefore. Now there will 10"

be three throws in all but the last round; in the last the expected number of throws is 2. Thus E(T) (1)x3 + 2 = 9. Finally, we remark that the method of Note 4 may be used to find the mean of the geometric distribution, as an alternative to the more usual method involving the summation of series. 1B.5 The randomised response experiment Let 0 denote the probability that a randomly sampled individual in some population voted Conservative in the last General Election. In a particular type of 'randomised response' experiment, each of a random sample of individuals from this population responds 'True' or 'False' to one of the following two statements. (a) I voted Conservative at the last General Election. (b) I did not vote Conservative at the last General Election. A randomising device is used to ensure that the probabilities of responding to (a) and (b) are p and 1-p respectively, where p is known and 0<p<1. If A is the probability that an individual responds 'True', write down an expression for A in terms of p and 0. For a group of mathematics teachers attending a statistics course, 24 out of 43 responded 'True' in an experiment in which p was fixed at 0.3. Use these figures to estimate 0.